Solved papers for JEE Main & Advanced AIEEE Solved Paper-2003

done AIEEE Solved Paper-2003 Total Questions - 3

• question_answer1) A circular disc X of radius R is made from an iron plate of thickness t and another disc Y of radius 4R is made from an iron plate of thickness t/4. Then, the relation between the moment of inertia \[{{I}_{X}}\] and \[{{I}_{Y}}\] is
  A)
  \[{{l}_{Y}}=32\,{{l}_{X}}\]                             
  done clear
  B)
  \[{{l}_{Y}}=16\,{{l}_{X}}\]                             
  done clear
  C)
  \[{{l}_{Y}}=\,{{l}_{X}}\]                 
  done clear
  D)
  \[{{l}_{Y}}=\,64{{l}_{X}}\]
  done clear
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• question_answer2) A particle performing uniform circular motion has angular momentum L. If its angular frequency is doubled and its kinetic energy halved, then the new angular momentum is     AIEEE  Solved  Paper-2003
  A)
  \[\frac{L}{4}\]                   
  done clear
  B)
        2L                           
  done clear
  C)
  4L                           
  done clear
  D)
  \[\frac{L}{2}\]
  done clear
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• question_answer3) Let F be the force acting on a particle having position vector r and \[\tau \] be the torque of this force about the origin. Then,     AIEEE  Solved  Paper-2003
  A)
  \[r\,.\,\tau =0\] and \[F\,.\,\,\tau \ne 0\]            
  done clear
  B)
  \[r\,.\,\,\tau \ne 0\] and \[F\,.\,\,\tau =0\]
  done clear
  C)
  \[r\,.\,\,\tau \ne 0\] and \[F\,.\,\,\tau \ne 0\]
  done clear
  D)
  \[r\,.\,\,\tau =0\] and \[F\,.\,\,\tau =0\]
  done clear
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